Question

Prove that \( \sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} \).

Hint:

When converting between inverse trigonometric functions, drawing a right-angled triangle can be a very intuitive and fast method. If \( \sin \theta = x/1 \), the opposite side is \(x\) and the hypotenuse is 1. The adjacent side will be \( \sqrt{1^2 - x^2} \), from which you can find any other trigonometric ratio.

Answer 1

Step 1: Understanding the Concept:
This question requires proving an identity involving inverse trigonometric functions.
The relationship between \( \sin^{-1} x \) and \( \cos^{-1} y \) can be established using a right-angled triangle and the fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
Step 2: Key Formula or Approach:
Let \( \theta = \sin^{-1} x \). This implies \( \sin \theta = x \).
We will use the identity \( \cos \theta = \sqrt{1 - \sin^2 \theta} \) (for the principal value range of \( \sin^{-1} x \), which is \( [-\pi/2, \pi/2] \), \( \cos \theta \) is non-negative).
From \( \cos \theta \), we can find \( \theta \) in terms of \( \cos^{-1} \).
Step 3: Detailed Explanation:
Let's start by assuming \( \theta = \sin^{-1} x \).
By the definition of the inverse sine function, this means: \[ \sin \theta = x \] Now, we use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
We can rearrange this to solve for \( \cos \theta \): \[ \cos^2 \theta = 1 - \sin^2 \theta \] Substituting \( \sin \theta = x \) into the equation: \[ \cos^2 \theta = 1 - x^2 \] Taking the square root of both sides gives: \[ \cos \theta = \sqrt{1 - x^2} \] (We take the positive root because the range of \( \sin^{-1} x \) is \( [-\pi/2, \pi/2] \), and in this interval, \( \cos \theta \geq 0 \)).
Now, we can express \( \theta \) using the inverse cosine function: \[ \theta = \cos^{-1} \sqrt{1 - x^2} \] Since we initially let \( \theta = \sin^{-1} x \), we can equate the two expressions for \( \theta \): \[ \sin^{-1} x = \cos^{-1} \sqrt{1 - x^2} \] Step 4: Final Answer:
We have successfully shown that both sides of the equation are equal to the same angle \( \theta \), thus proving the identity.